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FrozenGate by Avery

Pink Laser 500mW






Laser_Ben said:
The famous "Nuclear Flamingo Pink" you may read about is done with a 430nm blue and a 632nm HeNe in proper proportions. It is stunning, but nowhere near 500mW is possible. I was able to get 32mW out of my setup with my Coherent D3 430nm and a large Spectra Physics 124A HeNe
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1.Do this again.
2.Take pictures
3.????
4.PROFIT!

Seriously, you gotta do it again. You're the only one with that kind of equipment lying around! :D
 
Ah yeah, well pigments like paint work differently to light when mixed. Mix a bunch of paints and you get brown/grey/black etc. Mix a lot of different colored light beams and you get white. Sorry, you can't make a brown laser.
 
Your argument seems to be stating that when you destructively interfere with light, you create the color black The result is simply that the light is no longer visible. However, just like ocean waves passing through each other temporarily, if the light waves could be separated again, you would then be able to see them again. As you stated, the light isn't gone, the energy isn't gone, the photons themselves are not gone, they are simply a +1 and -1 state that overlaps to appear as a 0 state. but separate those +1 and -1 states again, and you no longer have a 0 state, but two separate +1 and -1 states. Given that the energy is still there, but you can not see it, I think the word invisible applies quite well, don't you?
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Ok lets define colour... colour is a visual attribute of things that results from the light they emit or transmit or reflect; "a white color is made up of many different wavelengths of light"

Now as we have both stated, "black" is not a colour.. it is a state in which there is an absence of light where it is totally absorbed/cancelled. When there is a +1 and a -1 and they are synced perfectly, there is a resulting of 0, or an absence of light, which we have already established is black.

You also keep bringing up other light rays or ambient light into the equation. I am purely speaking from a scientific point of view, and talking about a very specif and controlled situation (as stated many times before). In the ideal situation there is no ambient light, just direconal light.

Think of the example, you are in a completely dark room, and there are alot of dust particles in the air, then you turn on one flood light and it tottally fills the room with light, and you can see all the dust particles floating around. Now if you get a large ball bearing and throw it up then step back so you are viewing it from the side... what would you see? you are going to see a "black beam" because of the absence of light in that area. If you got a light meter and checked the intensity of that beam, it would equal 0.

Same thing is happening during deconstructive interference. The light waves at that point cancel creating a 0 spot, therefore creating, as we perseive a black beam or absence of light.

:D -Adrian
 
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bobhaha said:
Now as we have both stated, "black" is not a colour.. it is a state in which there is an absence of light where it is totally absorbed/cancelled. When there is a +1 and a -1 and they are synced perfectly, there is a resulting of 0, or an absence of light, which we have already established is black.
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Agreed, no argument there.



You also keep bringing up other light rays or ambient light into the equation. I am purely speaking from a scientific point of view, and talking about a very specif and controlled situation (as stated many times before). In the ideal situation there is no ambient light, just direconal light.

Think of the example, you are in a completely dark room, and there are alot of dust particles in the air, then you turn on one flood light and it tottally fills the room with light, and you can see all the dust particles floating around. Now if you get a large ball bearing and throw it up then step back so you are viewing it from the side... what would you see? you are going to see a "black beam" because of the absence of light in that area. If you got a light meter and checked the intensity of that beam, it would equal 0.
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Umm you can't have it both ways :D you can't rule out ambient light, as ANY hypothetical black beam would be in essence defined by that very ambient light. In your hypothetical room, with the flood light turned on, there is ambient light to cast the shadow you are calling a "black beam" however without the flood light on, you just have a dark room, and a ball bearing that you can't see. By the same logic, any darkness is defined by contrast with light. As you could not perceive a lack of light in the shape of a beam without light surrounding it to give it a boundary. It's the same logic as a hole isn't a hold if you don't have something there for it to be a hole IN :D


Lol this is quite entertaining, even if this is a bad case of threadjacking.
 
Umm you can't have it both ways you can't rule out ambient light, as ANY hypothetical black beam would be in essence defined by that very ambient light. In your hypothetical room, with the flood light turned on, there is ambient light to cast the shadow you are calling a "black beam" however without the flood light on, you just have a dark room, and a ball bearing that you can't see. By the same logic, any darkness is defined by contrast with light. As you could not perceive a lack of light in the shape of a beam without light surrounding it to give it a boundary. It's the same logic as a hole isn't a hold if you don't have something there for it to be a hole IN
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This is true but it was only an example, which I was tring to relate back to the path that the photons would travel to create the desconstructive interference (DI). So if the path is defined by the direction of the photon then that is its set boundry. I was also tring to relate the light meter readings to the DI cancellation of the light, ie +1 and -1. See what I'm getting at?

Lol this is quite entertaining, even if this is a bad case of threadjacking.
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hahaha totally agree... :D
 
bobhaha said:
Now as we have both stated, "black" is not a colour.. it is a state in which there is an absence of light where it is totally absorbed/cancelled. When there is a +1 and a -1 and they are synced perfectly, there is a resulting of 0, or an absence of light, which we have already established is black.

You also keep bringing up other light rays or ambient light into the equation. I am purely speaking from a scientific point of view, and talking about a very specif and controlled situation (as stated many times before). In the ideal situation there is no ambient light, just direconal light.

Think of the example, you are in a completely dark room, and there are alot of dust particles in the air, then you turn on one flood light and it tottally fills the room with light, and you can see all the dust particles floating around. Now if you get a large ball bearing and throw it up then step back so you are viewing it from the side... what would you see? you are going to see a "black beam" because of the absence of light in that area. If you got a light meter and checked the intensity of that beam, it would equal 0.

Same thing is happening during deconstructive interference. The light waves at that point cancel creating a 0 spot, therefore creating, as we perseive a black beam or absence of light.

:D -Adrian
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I would argue this isn't a laser in the traditional sense at all though. Sure you could manufacture a scenario wherein there is destructive interference with light, but that doesn't exactly create a black beam, it, as stridast has already pointed out, just cancels out light. The same principle stands with sound and how sound waves can constructively interfere and destructively interfere. Where the interfere destructively there is just an absence of the produced sound, not any other ambient sound.

Saying that it would be a controlled experiment with just directional light is nonsensical to me...A laser BEAM is ambient light, it's the photons of the laser getting diffracted by particles in the air, so wouldn't you ALWAYS have ambient light unless you had a perfect vacuum? And even in that scenario, in a perfect vacuum with just directional light interfering destructively, you would initially measure no light, then turn on the system and again measure an intensity of 0...So black becomes black...:P
 
I would argue this isn't a laser in the traditional sense at all though. Sure you could manufacture a scenario wherein there is destructive interference with light, but that doesn't exactly create a black beam, it, as stridast has already pointed out, just cancels out light. The same principle stands with sound and how sound waves can constructively interfere and destructively interfere. Where the interfere destructively there is just an absence of the produced sound, not any other ambient sound.

Saying that it would be a controlled experiment with just directional light is nonsensical to me...A laser BEAM is ambient light, it's the photons of the laser getting diffracted by particles in the air, so wouldn't you ALWAYS have ambient light unless you had a perfect vacuum? And even in that scenario, in a perfect vacuum with just directional light interfering destructively, you would initially measure no light, then turn on the system and again measure an intensity of 0...So black becomes black...:P
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Sorry but your not seeing the point...

absence of light = black
+1 -1 = 0
0 = absence of light
+1 - 1 = black

Therefore DI = black.. I really can't explain it any clearer than that because I don't have the vocabulary or clarity in English to do so. I understand in my head what I'm trying to say I'm afraid I just can express it for you to understand.

When absence of light is contrasted to light, you will see black as a colour... maybe this is where the confusion lies. Black is not a colour, it is merely absence of light. During DI there is an absence of light... If you read my other posts you will understand how this is done and I'm not going to explain it again. So since DI is absence of light, it must = black... therefore, to create a black laser, it must be done in extreme controlled conditions.

A laser is not ambient light. Ambient light can mean one of two things, either it is produced but heating an element where it becomes white hot producing light from electrical excitement of atoms and no photons are involved, or simply means background light "noise". A laser is a stimulated emition of photons, these photons are only responsible of colour when they have hit another atom creating light as a byproduct.

I don't think I can explain my self any better then I have so far... I will think about it and try again later..

-Adrian
 
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The point is kind of that black is the absence of all visible light (in the ultimate sense) So a laser beam in the DI conditions, wouldn't stop other light. the visible effect would only appear black in a black room, which wouldn't really accomplish anything you couldn't already accomplish by simply not turning a normal laser on in the first place. (hey, I guess I've got a black laser going in it's case in my desk right now! you wouldn't BELIEVE the battery life I am getting too!!!!!!!!!!) In a well lit environment, all you will have is invisible photons coming from the DI laser beam. all other light would be uneffected. You do understand this, you have already indicated as much. I think the misscommunication is this. We are talking in context with the original black laser posts in this thread, which describe seeing a black beam in broad daylight. You however at first said it was possible, but then continue to describe a situation that doesn't fit with the older posts :D Yes you could achieve a laser that is giving off photons that will no longer be visible to any normal means of detection, but are still technically there. But we are arguing that this doesn't fit the initial criteria of making that impossible visually black perceived "color" in sunlight. Yes black is technically the absence of color. But as our eyes, and brains, perceive it to be a color, and the earlier topic was describing a laser that would be a: visible in daylight, and b: appear to have a black beam, then to meet the criteria to claim the initial trolls, er I mean posts, were possible, you would have to come up with a different method :D
 
I'm not arguing that black is a color, I know what black is; it's our brains interpretation of what happens when our eyes don't send signals for a particular area in our field of view...So it's our perception of the absence of light.

When I say "ambient light" i just mean any light that is diffracted from it's source. This is the beam of a laser is, the photons from the laser being absorbed then re-emitted by the the atoms it encounters.

I reread your original post and agree with stridast's responses...Destructive interference of light waves is of course possible, as it is with any other waves. However this would not result in a black beam that a PERSON could distinguish. Sure it would result in an absence of light, but there is not way to distinguish that absence from the surroundings.

EDIT: Did not see stridast's response that was before mine lol he pretty much said the same thing as me.
 
The point is kind of that black is the absence of all visible light (in the ultimate sense) So a laser beam in the DI conditions, wouldn't stop other light. the visible effect would only appear black in a black room, which wouldn't really accomplish anything you couldn't already accomplish by simply not turning a normal laser on in the first place. (hey, I guess I've got a black laser going in it's case in my desk right now! you wouldn't BELIEVE the battery life I am getting too!!!!!!!!!!) In a well lit environment, all you will have is invisible photons coming from the DI laser beam. all other light would be uneffected. You do understand this, you have already indicated as much. I think the misscommunication is this. We are talking in context with the original black laser posts in this thread, which describe seeing a black beam in broad daylight. You however at first said it was possible, but then continue to describe a situation that doesn't fit with the older posts Yes you could achieve a laser that is giving off photons that will no longer be visible to any normal means of detection, but are still technically there. But we are arguing that this doesn't fit the initial criteria of making that impossible visually black perceived "color" in sunlight. Yes black is technically the absence of color. But as our eyes, and brains, perceive it to be a color, and the earlier topic was describing a laser that would be a: visible in daylight, and b: appear to have a black beam, then to meet the criteria to claim the initial trolls, er I mean posts, were possible, you would have to come up with a different method
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That's exactly right, expect I'm saying it is only possible in an extremely controlled environment, not out in the sun light, which i have said many times and proven through the evidence i have stated above.
 
:fightin:

epic discussion...

and lol @ ninjatux

peace & thanks
-cmak
 
mfo said:
..........

Do I need to draw this out on cue cards for you?

Black lasers are impossible.
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How do lcd's make black? I know that oled screens make black by simply turning off the pixel but I'm not sure about lcd's.
 
LCD screens are using the properties of the liquid crystal to twist light around perpendicular polarizations. the top part of an LCD screen is polarized horizontally. Then you have the liquid crystal layer, and the bottom part is polarized vertically. Beneath that you have a mirror. light that passes through without being twisted can't get through both polarization filters. Light that is being twisted around can. The electric current changes the structure of the liquid crystal, selectively controlling where the light is mirrored through, and where the light is blocked.

This is a very oversimplified explanation. "Twisting" is a poor choice of word to explain what is happening to the light, it just helps to get the concept across easier. Also, this describes LCD displays that are monochrome. i.e. a watch display or calculator display.
I assume for full color rendering, there is a backlight instead of a mirror, with tiny points in 3 different colors for each pixel. However that's an assumption, as the last time I read up on how a LCD screen worked, I was in junior high, and the original nintendo system was still around, and color LCD displays were pretty much nonexistent.
 
I have anticipated a name for the upcoming,... black laser !

Blackbox.;)
 
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